There are some techniques which assume a puzle has a single, unique solution. For example, if you find four squares which line up as the corners of a rectangle, and the possibilities are as follows:

Code:

AB . . ABC
. .
AB . . AB

Then the top right square must be <C>, if the puzzle is unique. If the value is not <C>, then the puzzle has multiple (or possibly zero) solutions.

You will see more when the solutions to these puzzles are posted. Also, if you search on the internet for "Sudoku Susser" you will find a very good PDF manual that explains this. Look at the sections on "Unique Rectangles" and "BUGs".

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The basic observation is, that if the puzzle has a unique solution, it cannot contain the pattern of possibilities

Code:

24 24
. .
24 24

So, one of the lower squares must be 6 or 8. Not very helpful (here). But, if you look at the row and block containing the lower squares, one of them must be 2. There are no other squares that have possibility 2. Therefore, neither of them can be possibility 4. So, remove 4 as possibilities in R8C1 and R8C2.

Then, after a few row/column/block interactions, you will reach the following:

As a relative newbie (two months), I could ask a dozen questions every day, but will try to limit myself, at least for the time being. I would like to ask if any action can be taken in the grid below. This is just one box from a puzzle and I wonder if anything can be done by looking at this box only, without considering the rest of the puzzle.

One of the things I wanted to know, but didn't specify, was whether anything could be done due to the fact that the pairs 13, 67 and 68 each appeared within three cells. I'm virtually certain that the answer is "no"; otherwise you would have noted it.

The David Bodycombe puzzles (published locally in the Detroit Free Press) have become significantly more interesting in the past couple of weeks.

Here is the puzzle from yesterday (5 stars)

Quote:

Both of these can be solved by assuming uniqueness. If you do not, you may have to find wings and forcing chains.

I tried that first puzzle, knowing that I wouldn't be able to solve it. But I DID solve it, much to my surprise. When I got down to about 20 unsolved cells, I hit a brick wall. Most of the cells held pairs, a few held triples. Finally, I got it by using my first-ever forcing chain.

However, it sounds like I wouldn't have needed a chain if I had "assumed uniqueness." That concept still puzzles (pun intended) me. I've been doing puzzles without assuming anything. Other than the treatment of the unique rectangles you illustrated in an earlier post, what would one do differently if he did or did not "assume uniqueness"?

The "constellation" lies in r1c4, r2c5, r7c5, r8c5, & r8c4. We can easily infer that r7c5 = 2 by noticing that r7c5 = 3 is impossible.

-- r7c5 = 3 ==> r8c5 = 7 ==> r8c4 = 5 ==> r1c4 = 9 ==> r2c5 = 7

But we can't have two "7"s in column 5, so r7c5 = 3 is impossible.

Someone_somewhere coined the term "5-star constellation" for this sort of formation. Basically it's a closed loop within the puzzle involving just 5 cells that can only exist in one state. Once you spot one it's easy to analyze. The tricky part is seeing the closed loop in the first place. dcb

It seems to me that the exclusion implied by the '5-star constellation' is really an xy-wing (which Keith noted in his post of Feb. 12). The cell r7c5 really has nothing to do with it. The fact that r8c5 can't be a 7 because that leaves r1c4 with no candidate is the key. Assigning R7c5 = 3 to start the loop merely adds an unnecessary step to find the contradiction. Otherwise we could say it was an '8-star galaxy' with r3c7, r3c9 and r7c9 as the added stars and that r3c7 can't be 2. They all lead to the fact that r8c5 can't be 7, which can be deduced from the three cell xy-wing.

I am not familiar with the definition and proof of a '5-star constellation' so maybe the fact that an xy-wing makes the same exclusion in this case is merely coincidental and that in a different but similar situation the '5-star constellation' could make the exclusion where something like an xy-wing would not work. If that is true, then I apologize in advance.

In short, r5c8 must contain either a "4" or a "5", and no matter which value goes there, r5c1 must be a "9". Unfortunately, that isn't enough to solve this particular puzzle. Suggestions are welcome! dcb

The BUG square is the one with possibilities <579>. Here is an explanation:

Quote:

In a BUG pattern, in each row, column and block, each unsolved possibility appears exactly twice. Such a pattern either has 0 or 2 solutions, so it cannot be part of a valid Sudoku.

When a puzzle contains a BUG, and only one square in the puzzle has more than 2 possibilities, the only way to kill the BUG is to remove both of the BUG possibilities from the square, thus solving it.

R2C6 - remove <57> from <579> leaving <9>.

Another way to solve this is via an XY-wing. Look at R1C4, then R8C4 and R2C5. One of R8C4 and R2C5 must be <7>, so R8C5 cannot be <7>, it must be <3>. The rest of the solution is simple.

Keith

Keith (or anyone),

I'm having a problem understanding the BUG pattern. It says "In a BUG pattern, in each row, column and block, each unsolved possibility appears exactly twice."

Then, "When a puzzle contains a BUG, and only one square in the puzzle has more than 2 possibilities, the only way to kill the BUG is to remove both of the BUG possibilities from the square, thus solving it.

R2C6 - remove <57> from <579> leaving <9>."

So we're "killing" something where two of the possibilities (5 & 7) occur twice and one thrice (9), even though it says in a BUG pattern each unsolved possibility appears exactly twice. I'm confused.

But regardless of my understanding, are we saying that when we have in one row, column or box the pattern XY-YZ-XYZ and no other cells have more than two possibilities, then we can remove XZ from the XYZ cell?

Somehow, I think it must be more complicated than that, but I've looked at that many times and must be missing something.

And finally, just curious, it appears that BUG is an acronym. What does it stand for?

First, BUG stands for "bivalue universal grave," which is sort of meaningless ... I think someone cooked it up just to get the cute acronym. The idea is an extension of the more familiar "non-unique rectangle" -- it (a BUG) is any pattern of unresolved cells that can be completed in more than one way.

Let's think about Keith's example above. If the possibilities at r2c6 were {5, 7} then the pattern of unresolved cells _for the whole puzzle_ would be a "BUG". We would either be able to complete it two different ways, or else we couldn't complete it (consistently) at all.

So the only way the pattern can define a unique solution is if r2c6 contains a "9". dcb

PS The "BUG" pattern doesn't just exist in a single row, or column. It exists throughout the entire puzzle. The ones I've run into occupied somewhere between 15 and 30 cells, roughly.